Topic
- Math
Description
https://leetcode.com/problems/divide-two-integers/
Given two integers dividend and divisor, divide two integers without using multiplication, division, and mod operator.
Return the quotient after dividing dividend by divisor.
The integer division should truncate toward zero, which means losing its fractional part. For example, truncate(8.345) = 8 and truncate(-2.7335) = -2.
Note:
- Assume we are dealing with an environment that could only store integers within the 32-bit signed integer range: [−2³¹, 2³¹ − 1]. For this problem, assume that your function returns 2³¹ − 1 when the division result overflows.
Example 1:
Input: dividend = 10, divisor = 3
Output: 3
Explanation: 10/3 = truncate(3.33333..) = 3.
Example 2:
Input: dividend = 7, divisor = -3
Output: -2
Explanation: 7/-3 = truncate(-2.33333..) = -2.
Example 3:
Input: dividend = 0, divisor = 1
Output: 0
Example 4:
Input: dividend = 1, divisor = 1
Output: 1
Constraints:
- -2³¹ <= dividend, divisor <= 2³¹ - 1
- divisor != 0
Analysis
方法一：我写的。
一开始，打算将被除数，除数绝对值后（先得出结果正负号），然后二分查找一个候选商，然后除数累加商值数，累加值与被除数相比，从而得出最佳候选商。但这方法在溢出情况处理很棘手，特别是被除数是Integer.MIN_VALUE情况（另外，Integer.MIN_VALUE == Math.abs(Integer.MIN_VALUE)为true，负数的绝对值还是负数，这情况是不允许的）。因此，放弃这种方法。
换种另一个角度，先得出结果正负号，让被除数，除数都是变成负数，然后让除数左移若干位，逼近被除数，逼近得差不多，得出半成商。接着就换成用加法逼近，得出最佳近似商。最后，根据一开始保留结果正负号，得出最佳近似商正负号，并将近似商返回。
方法二 & 方法三：别人写的，简洁，优雅。
Submission
public class DivideTwoIntegers {
//方法一：我写的
public int divide(int dividend, int divisor) {
if (divisor == 0)
throw new ArithmeticException();
if (dividend == Integer.MIN_VALUE && divisor == -1)
return Integer.MAX_VALUE;// 针对溢出情况
boolean negative = dividend > 0 ^ divisor > 0;
if (dividend > 0) dividend = -dividend;
if (divisor > 0) divisor = -divisor;
int quotient = 0, product = 0;
// 粗略得出商
for (int i = 0; i < 32; i++) {
int temp = divisor << i;// 相当于divisor * 2的i次方
// temp >> i != divisor, 溢出情况
if (temp < dividend || temp >> i != divisor) {
break;
} else if (temp > dividend) {
product = temp;
quotient = 1 << i;
} else {
return (negative ? -1 : 1) << i;
}
}
// 较精确逼近商
while (true) {
int temp = product + divisor;
// temp > 0, 溢出情况
if (temp < dividend || temp > 0)
break;
product = temp;
quotient++;
}
return negative ? -quotient : quotient;
}
//方法二：
public int divide2(int A, int B) {
if (A == Integer.MIN_VALUE && B == -1)
return Integer.MAX_VALUE;
int a = Math.abs(A), b = Math.abs(B), res = 0, x = 0;
while (a - b >= 0) {
for (x = 0; a - (b << x << 1) >= 0; x++)
;
res += (1 << x);
a -= (b << x);
}
return (A > 0) == (B > 0) ? res : -res;
}
//方法三：
public int divide3(int A, int B) {
if (A == Integer.MIN_VALUE && B == -1)
return Integer.MAX_VALUE;
int a = Math.abs(A), b = Math.abs(B), res = 0;
for (int x = 31; x >= 0; x--)
if ((a >>> x) - b >= 0) {
res += 1 << x;
a -= b << x;
}
return (A > 0) == (B > 0) ? res : -res;
}
}
Test
import static org.junit.Assert.*;
import org.junit.Test;
public class DivideTwoIntegersTest {
@Test
public void test() {
DivideTwoIntegers obj = new DivideTwoIntegers();
assertEquals(3, obj.divide(10, 3));
assertEquals(-2, obj.divide(7, -3));
assertEquals(0, obj.divide(0, 1));
assertEquals(1, obj.divide(1, 1));
assertEquals(Integer.MAX_VALUE, obj.divide(Integer.MAX_VALUE, 1));
assertEquals(Integer.MIN_VALUE, obj.divide(Integer.MIN_VALUE, 1));
assertEquals(0, obj.divide(1, Integer.MIN_VALUE));
assertEquals(-1, obj.divide(-1, 1));
assertEquals(1, obj.divide(Integer.MAX_VALUE, Integer.MAX_VALUE));
assertEquals(Integer.MAX_VALUE, obj.divide(Integer.MAX_VALUE, 1));
assertEquals(-Integer.MAX_VALUE, obj.divide(Integer.MAX_VALUE, -1));
assertEquals(-Integer.MAX_VALUE, obj.divide(-Integer.MAX_VALUE, 1));
assertEquals(Integer.MAX_VALUE, obj.divide(-Integer.MAX_VALUE, -1));
assertEquals(0, obj.divide(1, Integer.MAX_VALUE));
assertEquals(0, obj.divide(1, -Integer.MAX_VALUE));
for (int i = -100; i <= 100; i++) {
for (int j = 1; j < 100; j++) {
assertEquals(i / j, obj.divide(i, j));
}
}
}
@Test
public void test2() {
DivideTwoIntegers obj = new DivideTwoIntegers();
assertEquals(3, obj.divide2(10, 3));
assertEquals(-2, obj.divide2(7, -3));
assertEquals(0, obj.divide2(0, 1));
assertEquals(1, obj.divide2(1, 1));
assertEquals(Integer.MAX_VALUE, obj.divide2(Integer.MAX_VALUE, 1));
assertEquals(Integer.MIN_VALUE, obj.divide2(Integer.MIN_VALUE, 1));
assertEquals(0, obj.divide2(1, Integer.MIN_VALUE));
assertEquals(-1, obj.divide2(-1, 1));
assertEquals(1, obj.divide2(Integer.MAX_VALUE, Integer.MAX_VALUE));
assertEquals(Integer.MAX_VALUE, obj.divide2(Integer.MAX_VALUE, 1));
assertEquals(-Integer.MAX_VALUE, obj.divide2(Integer.MAX_VALUE, -1));
assertEquals(-Integer.MAX_VALUE, obj.divide2(-Integer.MAX_VALUE, 1));
assertEquals(Integer.MAX_VALUE, obj.divide2(-Integer.MAX_VALUE, -1));
assertEquals(0, obj.divide2(1, Integer.MAX_VALUE));
assertEquals(0, obj.divide2(1, -Integer.MAX_VALUE));
for (int i = -100; i <= 100; i++) {
for (int j = 1; j < 100; j++) {
assertEquals(i / j, obj.divide2(i, j));
}
}
}
@Test
public void test3() {
DivideTwoIntegers obj = new DivideTwoIntegers();
assertEquals(3, obj.divide3(10, 3));
assertEquals(-2, obj.divide3(7, -3));
assertEquals(0, obj.divide3(0, 1));
assertEquals(1, obj.divide3(1, 1));
assertEquals(Integer.MAX_VALUE, obj.divide3(Integer.MAX_VALUE, 1));
assertEquals(Integer.MIN_VALUE, obj.divide3(Integer.MIN_VALUE, 1));
assertEquals(0, obj.divide3(1, Integer.MIN_VALUE));
assertEquals(-1, obj.divide3(-1, 1));
assertEquals(1, obj.divide3(Integer.MAX_VALUE, Integer.MAX_VALUE));
assertEquals(Integer.MAX_VALUE, obj.divide3(Integer.MAX_VALUE, 1));
assertEquals(-Integer.MAX_VALUE, obj.divide3(Integer.MAX_VALUE, -1));
assertEquals(-Integer.MAX_VALUE, obj.divide3(-Integer.MAX_VALUE, 1));
assertEquals(Integer.MAX_VALUE, obj.divide3(-Integer.MAX_VALUE, -1));
assertEquals(0, obj.divide3(1, Integer.MAX_VALUE));
assertEquals(0, obj.divide3(1, -Integer.MAX_VALUE));
for (int i = -100; i <= 100; i++) {
for (int j = 1; j < 100; j++) {
assertEquals(i / j, obj.divide3(i, j));
}
}
}
}